Electrostatics Question 254
Question: A small sphere carrying a charge -q? is hanging in between two parallel plates by a string of length L. Time period of pendulum is $ T _{0} $ . When parallel plates are charged, the time period changes to $ T $ . The ratio $ T/T _{0} $ is equal to [UPSEAT 2003]
Options:
A) $ {{( \frac{pE}{I} )}^{1/2}} $
B) $ {{( \frac{pE}{I} )}^{3/2}} $
C) $ {{( \frac{I}{pE} )}^{1/2}} $
D) $ {{( \frac{p}{IE} )}^{1/2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
When dipole is given a small angular displacement q about it’s equilibrium position, the restoring torque will be
$ \tau =-pE\sin \theta =-pE\theta $ (as sinq = q) or $ I\frac{d^{2}\theta }{dt^{2}}=-pE\theta $ (as $ \tau =I\alpha =I\frac{d^{2}\theta }{dt^{2}} $ )
or $ \frac{d^{2}\theta }{dt^{2}}=-{{\omega }^{2}}\theta $
with $ {{\omega }^{2}}=\frac{pE}{I} $
therefore $ \omega =\sqrt{\frac{pE}{I}} $
